What is the extraneous solution to these equations? $\dfrac{x^2 + 15}{x - 8} = \dfrac{5x + 39}{x - 8}$
Solution: Multiply both sides by $x - 8$ $ \dfrac{x^2 + 15}{x - 8} (x - 8) = \dfrac{5x + 39}{x - 8} (x - 8)$ $ x^2 + 15 = 5x + 39$ Subtract $5x + 39$ from both sides: $ x^2 + 15 - (5x + 39) = 5x + 39 - (5x + 39)$ $ x^2 + 15 - 5x - 39 = 0$ $ x^2 - 24 - 5x = 0$ Factor the expression: $ (x - 8)(x + 3) = 0$ Therefore $x = 8$ or $x = -3$ At $x = 8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 8$, it is an extraneous solution.